Cardinality proof verification

Question

Problem:

Let $C \subset (0,1)$ be the set of all numbers whose unique decimal representation contains the number seven. Show that the number of elements in $C$ must be the same as the number of elements in $(0,1)$.

My solution:

We must construct a bijective function between $C$ and $(0,1)$ Let $(d_n)$ be the unique decimal representation of $x$.

Since $ \forall \space\space x\in C \space\space \exists! \space\space (d_n): x= (d_n) \space \implies f:C \to (0,1)\space\space s.t\space\space f(x) =x $ is injective.

Now we must show surjectivity, that is $ \forall x \in C \space\space f(x_1) \not = (fx_2) \implies x_1 \not = x_2$.

Let $f(x_1) = d_1$ and $f(x_2) = d_2$ since $d_n$ is unique $\implies f(x_1) \not = f(x_2) \implies (d_1) \not = (d_2) \implies x_1 \not = x_2 \forall x \in C$.

So the function is bijective and therefore the number of elements in C is the same as the number of elements in (0,1).

Any pointers? This is literally just a stab at this...I don't know if this is at all sufficient. Any advice would be great.

Answer 1

It's easy enough to make an injection either way: $C\subseteq (0,1)$ is the obvious injection one way, and for the other, there is a very simple bijection from $(0,1)$ to $(0.7,0.8)\subseteq C$.

Answer 2

Clearly $C$ is a proper subset of $(0,1)$, just as you noted in first words of the problem statement, so $$|C|\le |(0,1)|$$ OTOH, there is an injection $(0,1)\to C$, defined as follows:

• if a decimal expansion of a number contains sevens, replace every $7$ with $77$,
• otherwise, insert $7$ just after a decimal point

which implies $$|(0,1)| \le |C|$$ Now the Schröder–Bernstein theorem states that the two sets are equipollent: $$|C| = |(0,1)|$$